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Help.............Triangle ABC is a right triangle with angle ACB as the right angle. Angle ABC =60 degrees and AB=10. Let P be randomly chosen inside triangle ABC and extend BP to meet AC at D. What is the probability that BD>5*sqrt(2) Help.............Triangle ABC is a right triangle with angle ACB as the right angle. Angle ABC =60 degrees and AB=10. Let P be randomly chosen inside triangle ABC and extend BP to meet AC at D. What is the probability that BD>5*sqrt(2) @Mathematics
the answer must be 1/sqrt(3)
answer |dw:1320476924776:dw|
how did u take as 15 and 45?
i never said that CBD is a right isosceles triangle
since we want hat BD> 5sqrt(2)........(I) we know that....10/sine(BDA) = BD/ sine(DAB) since.... sine(DAB) = 1/2 therefore..... BD = 5/ sine(BDA) put in equation no....(I) we get..... 5/sine(BDA)> 5*sqrt(2) yaha tak smjh me aa gaya kya sb kuch????
so we get............ sin (BAD)<1/sqrt(2) or 180degre< BAD<135degree got it or not????
From here i understood how to solve Thankx very very much fr ure help.......
now do u understand the figure that i have made....its the boundary line of the probability...that means in that figure if we move d to the right then we will get the desired probability....... understood????
And onee day hopefully like u i will be an IITIAN Yeah i understood THANX A LOT.......
listen tio the important fact........ur answer would be now....... area of triangle(CBD)/ area oof the triangle (DBA) this is the main thing....
bye.....i'm leaving now...mujhe khana khane jana h aur bhi kuch kam h....i'll be back in two hrs now...do send me a request on facebook.......